hdu 4734 F(x) 数位DP

[align=left]Problem Description[/align]
For a decimal number x with n digits (AnAn-1An-2 ... A2A1), we define its weight as F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2
* 2 + A1 * 1. Now you are given two numbers A and B, please calculate how many numbers are there between 0 and B, inclusive, whose weight is no more than F(A).
 

[align=left]Input[/align]
The first line has a number T (T <= 10000) , indicating the number of test cases.

For each test case, there are two numbers A and B (0 <= A,B < 109)
 

[align=left]Output[/align]
For every case,you should output "Case #t: " at first, without quotes. The
t is the case number starting from 1. Then output the answer.
 

[align=left]Sample Input[/align]

3
0 100
1 10
5 100

 

[align=left]Sample Output[/align]

Case #1: 1
Case #2: 2
Case #3: 13

 题意：给定A和B,求给定区间1~B中的x对应的f(x)小于f（A）的数。
F(x) = An * 2n-1 + An-1 * 2n-2 + ... + A2
* 2 + A1 * 1;
思路：简单的数位DP+记忆化搜索，唯一注意的一点是初始化DP的时候要放在外面防止超时;
代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int dp[15][50000],b[15];
int f(int n)
{
int ans=0,t=1;
while(n)
{
ans+=(n%10)*t;
t=t*2;
n=n/10;
}
return ans;
}
int dfs(int len,int ans,int flag)
{
if(len<=0&&ans>=0)
return 1;
if(ans<0)
return 0;
if(!flag&&dp[len][ans]!=-1)
return dp[len][ans];
int end=flag?b[len]:9;
int s=0;
for(int i=0;i<=end;i++)
{
s=s+dfs(len-1,ans-i*(1<<(len-1)),flag&&i==end);
}
if(!flag)
dp[len][ans]=s;
return s;
}
int find(int n,int m)
{
int len=0;
while(m)
{
b[++len]=m%10;
m=m/10;
}
return dfs(len,f(n),1);
}
int main()
{
int t;
scanf("%d",&t);
memset(dp,-1,sizeof(dp));  //注意放在外面
for(int T=1;T<=t;T++)
{
int n,m;
scanf("%d%d",&n,&m);
printf("Case #%d: ",T);
printf("%d\n",find(n,m));
}
}